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Maslowich
2 years ago
7

During the day, susan notices that the wind is blowing onshore at the beach. What is this called? land breeze land breeze sea br

eeze sea breeze monsoon monsoon global wind
Physics
1 answer:
In-s [12.5K]2 years ago
8 0

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

  • Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
  • Sea breezes form as a result of changes in air pressure brought on by the different heat capacities of water and dry land. Sea breezes are therefore more confined than prevailing winds.
  • A sea wind is frequently seen along coasts after sunrise because land warms up far more quickly than water does when exposed to solar radiation.
  • The sea wind front is significant because it can serve as a catalyst for afternoon thunderstorms and provide welcome cooling along the coast.

Learn more about the sea breeze with the help of the given link:

brainly.com/question/13015619

#SPJ4

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Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
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3 years ago
PLEASE HELP I'LL GIVE BRAINLIEST: Consider a sound wave produced by a tuning fork. Through which materials would the sound trave
ra1l [238]

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An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life
elena55 [62]

Answer:

The half-life is t_{1/2} = 1.005 h

Explanation:

Using the decay equation we have:

A=A_{0}e^{-\lambda t}

Where:

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  • A(0) the initial activity
  • A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that A = \frac{A_{0}}{2}

\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}

0.5=e^{-\lambda*1 h}

Taking the natural logarithm on each side we have:

ln(0.5)=-\lambda

\lambda=0.69 h^{-1}

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

\lambda = \frac{ln(2)}{t_{1/2}}

t_{1/2} = \frac{ln(2)}{\lambda}

t_{1/2} = \frac{ln(2)}{0.69}

t_{1/2} = 1.005 h

I hope it helps you!

6 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
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To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

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The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

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4 0
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in a series circuit, how does the voltage supplied by the battery compare to the voltages on each load?
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Answer:

In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.

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<em>carry on</em><em> </em><em>learning</em><em> </em>

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